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Math Help - Find function base on first order derivatives

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    Find function base on first order derivatives

    Suppose that the function f: \mathbb {R}^2 \rightarrow \mathbb {R} has the properties of:

    a) f(0,0)=1
    b)  \frac { \partial f }{ \partial x } (x,y) = 2 and  \frac { \partial f }{ \partial y } (x,y) = 3 \ \ \ \ \ \forall (x,y) \in \mathbb {R}^2

    Prove that f(x,y)=1+2x+3y

    Well, I guess I can integrate the derivatives and get this answer, but I have to solve it with methods of mean value theorem and gradients... How should I proceed? Thank you.
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    Quote Originally Posted by tttcomrader View Post
    Suppose that the function f: \mathbb {R}^2 \rightarrow \mathbb {R} has the properties of:

    a) f(0,0)=1
    b)  \frac { \partial f }{ \partial x } (x,y) = 2 and  \frac { \partial f }{ \partial y } (x,y) = 3 \ \ \ \ \ \forall (x,y) \in \mathbb {R}^2

    Prove that f(x,y)=1+2x+3y

    Well, I guess I can integrate the derivatives and get this answer, but I have to solve it with methods of mean value theorem and gradients... How should I proceed? Thank you.
    Start by looking at the function g(x) = f(x,0). Apply the mean value theorem to show that g(x) = 1 + 2x. Then, knowing that f(x,0) = 1+2x, fix x and apply the mean value theorem to the function h(y) = f(x,y) to show that h(y) = h(0) + 3y = 1+2x+3y.
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